TITLE: A generalization of Liouville's theorem
QUESTION [5 upvotes]: Let $A\subseteq \Bbb{C}$ be countable and closed. Let $f: \Bbb{C}\backslash A \longrightarrow \Bbb{C}$ be analytic and bounded. Is it true that $f$ must be constant ?

REPLY [3 votes]: The proof proposed by edo arad is quite nice.
Here is another one, which is essentially the same but avoids the use of transfinite induction.
By Liouville, it is enough to show that $f$ can be extended to an analytic function on $\mathbb C$. 
To do this, it is enough to show that every point $a\in A$ has an open neighbourhood $V_a$ such that $f$ can be extended to an analytic function $f_a$ on $(\mathbb C\setminus A)\cup V_a$. Indeed, assume that this holds true. Then the $V_a$'s can be taken to be open disks, so that $V_a\cup V_{a'}$ is connected whenever $V_a\cap V_{a'}\neq\emptyset$. By the identity principle, it follows that if $V_a\cap V_{a'}\neq\emptyset$, then $f_a\equiv f_{a'}$ on $V_a\cap V_{a'}$: this is because $f_a$ and $f_{a'}$ agree on $V_a\cap V_{a'}\cap (\mathbb C\setminus A)$, which is a nonempty open set because $\mathbb C\setminus A$ is dense in $\mathbb C$ (recall that $A$ is countable, so it has empty interior). Therefore, there is a perfectly defined function $\widetilde f:\mathbb C\to\mathbb C$ such that $\widetilde f\equiv f$ on $\mathbb C\setminus A$ and $\widetilde f(z)\equiv f_a(z)$ on $V_a$ for any $a\in A$; and $\widetilde f$ is analytic because it is analytic in a neighbourhood of each point.
Let us denote by $S$ the set of all points $a\in S$ such that $f$ cannot be analytically extended in a neighbourhood of $a$. We have to show that $S=\emptyset$; and since $S$ is countable and obviously closed, it is enough to show that $S$ has no isolated points (because a nonempty closed subset of $\mathbb C$ without isolated points has to be uncountable, for example by the Baire category theorem).
Towards a contradiction, assume that $S$ has at least one isolated point $a$. This means that one can find an open neighbourhood $V$ of $a$ such that $V\cap S=\{ a\}$. In other words, $f$ can be analytically extended in a neighbourhood of every point $a'\in V\setminus\{ a\}$. By the above reasoning, $f_{V\setminus A}$ can be extended to an analytic function $g$ defined on $V\setminus\{ a\}$. This function $g$ is bounded on $V\setminus\{ a\}$ because $f$ is bounded on $V\setminus A$ and $V\setminus A$ is dense in $V$. But this implies that $g$ has a removable singularity at $a$. So $f_{V\setminus A}$ can in fact be extended to an analytic function on $V$, which contradicts the assumption $a\in S$.